The circuit in this problem has an ideal op-amp, two resistors, and one inductor.
The input signal is a sinusoidal voltage source.
We are interested in the output voltage in sinusoidal steady-state.
We can try to describe the circuit with a transfer function.
The transfer function is the s domain ratio of the output voltage to the input voltage.
We know the input signal and are interested in the output voltage.
We know the transfer function and input voltage frequency is ω1.
Here, we can try to evaluate the transfer function.
Just substitute s = jω1 into the transfer function.
The transfer function can be written as a complex number with a magnitude and phase
angle.
The output voltage magnitude is exactly the product of the transfer function magnitude
with the input signal magnitude.
We can add the phase angle of the input to that of the transfer function to get the phase
angle of the output.
First, let’s try to find the transfer function of the circuit.
We need to convert the circuit into s domain.
The impedance of an inductor is s L.
The Laplace transform of the input voltage is Vin.
We use upper case V to represent voltage in the s domain.
We need to relate Vin to Vout.
The two elements here can be combined into one impedance.
We can write a Kirchhoff’s Current Law equation at the inverting node to relate the input
voltage with the output voltage.
At the inverting node, the voltage is 0.
The current flow through Zin should be Vin – 0 over Zin.
The voltage divided by impedance is the current.
The current flow through Zf should be Vout – 0 over Zf.
No current flows into the ideal op-amp, so it’s 0 Amps here.
We have a voltage ratio, which is –Zf over Zin.
We can substitute Zf and Zin into the equation.
Zf is just a 10 kΩ resistor.
At Zin we have two impedances in series, so it’s 5 kΩ + 0.05 S.
We can simplify the transfer function by making the coefficient of s in the denominator
equal to 1.
The numerator and denominator are divided by 0.05.
[equation]
We have a transfer function in the s domain.
We are interested in the output voltage in sinusoidal steady-state.
Here we need to evaluate the transfer function at the frequency ω1.
ω1 in this problem is 100,000 rad/s.
We substitute j 100,000 for s into the transfer function.
The numerator is constant.
This is a complex number.
It can be written as a magnitude.
We get 1.414 at a phase angle of 135°.
We have the phase angle and the magnitude of the transfer function, so we can write the
output voltage in sinusoidal steady-state.
That is the magnitude of the transfer function multiplied by the input voltage magnitude
times cosine of the same frequency as the input frequency plus the phase angle of the
input signal plus the phase angle of the transfer function.
[final equation]
The unit is Volts.